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#1
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On 6/25/2010 8:37 PM, tom wrote:
On 6/25/2010 12:06 PM, Szczepan Bialek wrote: They must: " In electronic circuit theory, a "ground" is usually idealized as an infinite source or sink for charge, which can absorb an unlimited amount of current without changing its potential." Your mistake here is to assume that because they can sink current that the are sinking DC current. The "ground", as you like to think about it may carry current during the RF cycle, but there is nothing that would cause an unbalanced charge buildup since it IS symmetrical. tom K0TAR And I forgot to include that there is no field emission, so no charge could possibly build up. tom K0TAR |
#2
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![]() "tom" wrote . net... And I forgot to include that there is no field emission, so no charge could possibly build up. Emission is the result of voltage. If the power is ON the emissin take place. S* |
#3
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On 6/26/2010 2:38 AM, Szczepan Bialek wrote:
wrote . net... And I forgot to include that there is no field emission, so no charge could possibly build up. Emission is the result of voltage. If the power is ON the emissin take place. S* Incorrect. tom K0TAR |
#4
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![]() "tom" wrote . net... On 6/26/2010 2:38 AM, Szczepan Bialek wrote: wrote . net... And I forgot to include that there is no field emission, so no charge could possibly build up. Emission is the result of voltage. If the power is ON the emissin take place. Incorrect. "(II) Field emission: In this method, a strong electric field (i.e. a high positive voltage) is applied at the metal surface which pulls the free electrons out of the metal because of the attraction of positive field. The strong the electric field, the greater is the electron emission. " You are incorrect. S* |
#5
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On Jun 27, 9:28*am, "Szczepan Bialek" wrote:
*"tom" ouse.net... On 6/26/2010 2:38 AM, Szczepan Bialek wrote: *wrote e.net... And I forgot to include that there is no field emission, so no charge could possibly build up. Emission is the result of voltage. If the power is ON the emissin take place. Incorrect. "(II) Field emission: In this method, a strong electric field (i.e. a high positive voltage) is applied at the metal surface which pulls the free electrons out of the metal because of the attraction of positive field. The strong the electric field, the greater is the electron emission. " You are incorrect. S* that quote is correct... unfortunately the voltages created by most radio transmitters aren't high enough to overcome the metal work function so no field emission takes place from antennas. and even if it did it wouldn't go anywhere because radio waves are symmetric. |
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